PROP. II. PROB. - In a given circle to inscribe a triangle equiangular to a given triangle.

Let ABC be a given circle, and DEF a given triangle; it is required to inscribe in ABC a triangle equiangular to DEF.

Draw the straight line GAH (III. 17) touching the circle in any point A, and make (I. 23) the angle HAC equal to E, and

GAB equal to F, and join BC: ABC

G

is the triangle required.

A

For, since GAH touches the circle ABC, and AC is drawn from the point of contact, the angle HAC is equal (III. 32) to the angle B in the alternate segment; but HAC is equal (const.) to

E: therefore also B is equal to E. For the same reason, the angle C is equal to F: therefore the remaining angle BAC is equal (I. 32, cor. 5) to the remaining angle D: wherefore the triangle ABC is equiangular to the triangle DEF, and it is inscribed (IV. def. 2) in the circle ABC: which was to be done.

PROP. III. PROB.-About a given circle to describe a triangle equiangular to a given triangle.

Let ABC be

a

given circle, and DEF a given triangle; it is required to describe a a triangle about ABC, equiangular to DEF. Produce EF both ways to G, H, and find (III. 1) K the centre of the circle; from it draw any radius KB; and make (I. 23) the angle BKA equal to DEG, and BKC equal to DFH; through the points A, B, C, draw (III. 17) the tangents LAM, MBN, NCL: LMN is the required triangle.*

For, since LM, MN, NL touch the circle ABC in the points A, B, C, to which KA, KB, KC are drawn from the centre, the angles at the points A, B, C, are (III. 18) right angles: les: and because the four angles of the quadrilateral figure

A

D

L

C

GE

FH

K

M B

N

AMBK are equal (I. 32, cor. 1) to four right angles; and that two of them, KAM, KBM, are right angles, the other two AKB, AMB are equal to two right angles. But (I. 13) the angles DEG, DEF are likewise equal to two right angles: therefore the angles AKB, AMB are equal to the angles DEG, DEF, of which (const.) AKB is equal to DEG: wherefore the remaining angles AMB, DEF are equal. In like manner, the angle LNM may be demonstrated to be equal to DFE; and therefore the remaining angle MLN is equal (I. 32, cor. 5) to the remaining angle EDF: wherefore the triangle LMN is equiangular to DEF: and (IV. def. 3) it is described about the circle ABC: which was to be done.

PROP. IV. PROB. - To inscribe a circle in a given triangle.

Let ABC be a given triangle; it is required to inscribe a circle in it.

Bisect (I. 9) the angles ABC, BCA by the straight lines BD, CD, meeting one another in the point D; and from D draw (I. 12) DE, DF, DG, perpendiculars to AB, BC, CA. Then, because the angle EBD is equal (const.) to FBD, and that the right angles BED, BFD are equal, the two triangles EBD, FBD have two angles of the one equal to two angles of the other, each to each, and the side BD, which is opposite

equal to

G

E

D

B

F

to one of the equal angles in each, is common to both; therefore (I. 26) DE is equal to DF. In the same manner it would be shown, that, in the triangles DGC, DFC, DG is is equ DF: therefore the three straight lines DE, DF, DG are equal to one another, and the circle described from the centre D, at the distance of any of them, will pass through the extremities of the other two; and (III. 16, cor.) it will touch the straight lines AB, BC, CA, because the angles at the points E, F, G are right angles. Therefore the straight lines AB, BC, CA touch the circle, and the circle EFG is inscribed (IV. def. 3) in the triangle ABC; which was to be done.

to

Schol. This proposition is a case of the general problem, describe a circle touching three given straight lines which do not pass through the same point, and which are not all parallel to one another. If two of the lines be parallel, there may evidently be two equal circles, one on each side of the line falling on the parallels, each of which will touch the three given lines: and their centres will be the intersections of the lines bisecting the angles made by the parallels with the third line.

If the lines form a triangle by their intersections, there will be

four circles touching them;

scribed, and the others

literally, inscribed,
each touching one side externally,
and the other two produced. The
centres of the external circles will
be the intersections of the lines
bisecting two exterior angles; and
the line bisecting the remote inte-
rior angle will pass through the
same point. The method of proof
respecting the external circles is
almost the same as that of the
foregoing proposition.

Cor. 1. Join AD. Then (I. 47, cor. 5) the angles EAD, GAD are equal. Hence the three straight lines bisecting the three angles of a triangle meet in the same point, and that point is the centre of the inscribed circle.

Cor. 2. AE is equal (III. 17, schol.) to AG, BE to BF, and CF to CG. Hence the sum of any side AB, and the remote segment, ual to half the peri

CF or CG, of either of the other sides, is equal

meter: and consequently either of these segments, CF or CG, is

equal to what remains, when the

third side AB is taken from half

Cor. 4. Since (cor. 2) AB and FC are equal to half the perimeter, and AK equal (cor. 3) to the same; if AB be taken away there remains BK or BH equal to CF or CG.

Cor. 5. If C be a right angle, CG or CF is evidently equal to DG or DF, the radius of the inscribed circle. Hence, in a rightangled triangle the diameter of the inscribed circle is equal to the excess of the sum of the legs above the hypotenuse. It would appear, in like manner, that the diameter of the circle touching the hypotenuse externally, and the legs produced, is equal to the perimeter.

Cor. 6. Since the triangles ADB, BDC, and CDA are respectively equal (I. 41) to the rectangles under the radius of the inscribed circle, and the halves of AB, BC, and CA, it follows (II. 1) that the area of ABC is equal to the rectangle under the radius and half the perimeter. Hence, if the sides be given in numbers, the length of the radius may be computed by calculating (II. 13, note) the area, and dividing it by half the sum of the sides, or its double by their sum.

Cor. 7. If straight lines be drawn from M, the centre of one of the external circles, to A, B, and C, the triangles AMB, AMC, and BMC, are respectively equal (I. 41) to the rectangles under the radius of that circle and the halves of AB, AC, and BC. Hence, if the last of these be taken from the sum of the others, there remains the triangle ABC equal to the rectangle under the radius, and the excess of half the sum of AB and AC above the half of BC; or, which is the same, to the rectangle under the radius, and the excess* of half the perimeter above BC. The radius, therefore, of any of the external circles may be computed by dividing the area by the excess of half the perimeter above the side which the circle touches externally.

PROP. V. PROB. - To describe a circle about a given triangle.

Let the given triangle be ABC; it is required to describe a circle about it.

Find (III. 1, schol.) D the centre of the circle passing through A, B, C, the three angular points of the triangle, and the circle described from Das centre, at the distance of any of the angular points A, B, C, is evidently the circle required.

B

D

C

G

Cor. 1. It is evident (from this proposition, and from III. 1, schol.), that the three straight lines bisecting the three sides and perpendicular to them, meet in the same point, and that the point in which they meet is the centre of the circumscribed circle.

Cor. 2. Through the centre D draw DG perpendicular to BC: the angle BDG or CDG is equal to BAC. For (I. 4) the angles BDG, CDG are equal, as are also (III. 26) the arcs BG, CG. Therefore (III. 27) the straight line joining AG bisects the angle BAC: and (III. 20) each of the angles BDG, CDG is double of either half of BAC.

PROP. A. THEOR. * - Any equilateral and equiangular rectilineal figure may have one circle described about it, and another in it; and the same point is the centre of both circles.

Let ABCDE be a rectilineal figure, having all its sides equal, and all its angles equal: one circle may be described about it, and another in it, and the same point is the centre of both.

Bisect the angles BCD, CDE by the lines CF and DF: these (I. ax. 12) will meet, if produced: let them be produced and meet in F, and join FB. Then, since the angles FCD, FDC are equal, being the halves of equal angles, FC is equal

B

G

A

E

F

(I. 6) to FD. Also, in the triangles BCF, DCF, the side BC is equal (hyp.) to DC, CF is common, and (const.) the angle BCF is equal to DCF: therefore (I. 4) BF is equal to DF, and consequently to CF. The angle CBF also is equal to CDF, and is therefore the half of ABC. In the same manner, the straight lines drawn from F to the other angles A and E may be proved to be equal to BF, CF, or DF: and therefore a circle described from Fas centre, at a distance equal to any of these lines, will pass through all the angular points, and (IV. def. 2) will be circumscribed about the rectilineal figure ABCDE.

From F draw FG perpendicular to BC, and FH to CD. Then, since BC and CD are equal chords of the circumscribed circle, they are (III. 14) equally distant from the centre: that is, FG is equal to FH: and a circle described from Fas centre at the distance of either of these, will pass through G and H; and it will touch (III. 16) BC and CD, because the angles at Gand H are right angles: and in the same manner it might be shown, that the same circle would touch the other sides of the figure: it would therefore (IV. def. 3) be inscribed in the rectilineal figure ABCDE: wherefore any equilateral and equiangular rectilineal figure, &c.

Cor. BC is bisected (III. 3) in G, and CD in H; and (const.) FG is perpendicular to BC, and FH to CD. Hence the centre of the inscribed or circumscribed circle may be found by assuming any three of the angular points, and following the method pointed out in the scholium to the first proposition of the third book; or by bisecting two adjacent sides by perpendiculars: and the radius of the inscribed circle is the perpendicular drawn from F to any of the sides, while that of the circumscribed circle, is the straight line drawn from the same point to any of the angles.

PROP. B. THEOR. If any equilateral and equiangular rectilineal figure be inscribed in a circle, tangents to the circle, drawn through the angular points, will form an equilateral and equiangular figure of the same number of sides, described about the circle.

Let ABCDE be an equilateral and equiangular rectilineal figure described in the circle ACE: if tangents be drawn to the circle through the points A, B, C, D, E, meeting in the points F, G, H, K, L, the figure FGHKL is also an equilateral and equiangular figure of the same number of sides as the figure ABCDE.

For, since GH touches the circle in C, and CD cuts it, the

angle DCH is equal (III. 32) to any angle at the circumference standing on the arc CD; and because HK touches the circle in D, and CD cuts it, the angle CDH is also equal to any angle at the circumference standing on the same arc CD : therefore the angles DCH, CDH are equal to one another. For the same reason EDK, DEK are each equal to any angle at the circumference standing on the arc DE: but (III. 28) the arcs AB, BC,

CD, &c. are all equal, because (hyp.) their chords are equal; and (III. 27) angles at the circumference standing on these arcs are equal; and therefore the angles DCH, CDH, EDK, DEK, LEA, &c. are all equal. Hence (I. 6) CH is equal to HD, DK to KE, &c. Also, in the triangles CHD, DKE, there are two angles of the one equal to two angles of the other, each to each, and (hyp.) the sides CD, DE are equal: therefore (I. 26) the sides CH, HD are equal to DK, KE, each to each, and the angle H to K. In the same manner, it would be shown that the angle K is equal to